Answer:
+ 291.9 kJ
Solution:
The equation given is as;
H₂O ₍s₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = ?
First, as we know the heat of formation of H₂O ₍l₎ is,
H₂ ₍g₎ + 1/2 O₂ ₍g₎ → H₂O ₍l₎ ΔH = - 285.9 kJ
Now, reversing the equation will reverse the sign of heat as,
H₂O ₍l₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = + 285.9 kJ
Also, we know that,
H₂O ₍s₎ → H₂O ₍l) ΔH = + 6.0 kJ
Now, adding last two equations,
H₂O ₍l₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = + 285.9 kJ
H₂O ₍s₎ → H₂O ₍l) ΔH = + 6.0 kJ
-----------------------------------------------------------------------------
H₂O ₍s₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = + 291.9 kJ
Answer:
9.17 g
Explanation:
1) Calculate mols of PCl5
PCl5 (aq) + 4H2O (l) ⟶ H3PO4 (aq) + 5HCl (aq)
Mass of PCl5: 26.5g
Molar Mass of PCl5 208.24g/mol
Mol of PCl5 = Mass PCl5 /Molar mass PCl5 = 26.5g / (208.24g/mol) = 0.127257011 mol
2) Calculate mols of water needed to react
Mols of H2O per 1 Mols of PCl5 = 4 (because 4 water is needed to Phosphorus pentachloride).
Mols of H2O = Mol of PCl5 * 4 = 0.127257011 * 4 = 0.509028044 mols
3) Calculate Grams of water needed to react
Mols of H2O = 0.509028044 mols
Molar Mass of H2O = 18.015 g/mol
Mass of H2O = Mols of H2O * Molar Mass of H2O
= 0.509028044 mols* 18.015 g/mol = 9.17014021 g
sf = 3
9.17 g
